I'm trying to show the following.
Let $A\subseteq\mathbb{R}$. $\forall n\in\mathbb{N}\ (A\cap(n,n+1]\in\mathcal{B}((n,n+1])) \iff A\in\mathcal{B}(\mathbb{R})$,
where $\mathcal{B}$ is the Borel $\sigma$-algebra on the corresponding underlying set. I can show $\Rightarrow$ direction: Let $A_n=A\cap (n,n+1]$. Then $A=\cup_{n\in\mathbb{Z}}A_n$. It can be seen that $\mathcal{B}((n,n_1])\subseteq \mathcal{B}(\mathbb{R})$ since $\mathcal{B}(\mathbb{R})$ include all the open intervals within $(n,n+1]$ for every $n\in\mathbb{Z}$. Since $A_n\in\mathcal{B}((n,n_1])$, we have $A_n\in\mathcal{B}((n,n+1])$ and hence $A\in\mathcal{B}((n,n+1])$.
However, I don't know how to show the other direction.
Let $\mathcal{F}:=\{A\subseteq \mathbb{R} \mid A\cap(n,n+1]\in \mathcal{B}((n,n+1])\}$. It is clear that $\mathcal{F}$ is a $\sigma$-algebra of $\mathbb{R}$. As every open set of $\mathbb{R}$ lies in $\mathcal{F}$, we have $\mathcal{B}(\mathbb{R})\subseteq \mathcal{F}$. Thus $A\in \mathcal{B}(\mathbb{R})\implies A\cap(n,n+1]\in \mathcal{B}((n,n+1])$.